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Stoichiometric Calculations

1.5 g of chal...

Question

$1.5$ g of chalk was treated with $10$ mL of $4 N$ $H C l$ . The chalk was dissolved and the solution was made to $100$ mL. $25$ mL of this solution required $18.75$ mL of $0.2 N$ $N a O H$ solution for complete neutralisation. Calculate the percentage of pure $C a C O_{3}$ in the sample of chalk?

17

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67

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50

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None of the above

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Solution

The correct option is D $50$
$C a C O_{3} + 2 H C l \to C a C l_{2} + H_{2} C O_{3}$

$H_{2} C O_{3} + 2 N a O H \to N a_{2} C O_{3} + 2 H_{2} O$
Let the percentage of pure $C a C O_{3}$ in the sample of chalk be $X$ %.
1.5 g of chalk contains $\frac{1.5 X}{100} = 0.015 X$ g of
$C a C O_{3}$ . The molar mass of $C a C O_{3}$ is $100 g / m o l$ . The number of moles of $C a C O_{3} = \frac{0.015 X g}{100 g / m o l} = 1.5 X \times 10^{- 4}$ moles. Out of 100 mL of solution, 25 mL is taken. The number of moles of $H_{2} C O_{3}$ in 25 mL of solution $= \frac{1.5 X \times 10^{- 4}}{4} = 3.75 \times 10^{- 5} X$ .
Number of moles of $N a O H$ used $= \frac{18.75 m L}{1000 m L / L} \times 0.2 = 0.00375$ moles
Number of moles of $H_{2} C O_{3} = \frac{1}{2} \times$ Number of moles of $N a O H$
$3.75 \times 10^{- 5} X = \frac{1}{2} \times 0.00375$
$X = 50$ %
Hence, the percentage of pure $C a C O_{3}$ in the sample of chalk is 50%.

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1.5 g of chalk was treated with 10 mL of 4 N HCl. The chalk was dissolved and the solution was made to 100 mL. 25 mL of this solution required 18.75 mL of 0.2 N NaOH solution for complete neutralisation. Calculate the percentage of pure CaCO3 in the sample of chalk?

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